Asked by rebs
find the vertices, foci, equations of the asymptotes, length of LR, and the graph of the hyperbola x^2-4y^2+4x+24y=28
Answers
Answered by
Steve
in standard form, you have
x^2+4x - 4(y^2-6y) = 28
x^2+4x+4 - 4(y^2-6y+9) = 28+4-4*9
(x+2)^2 - 4(y-3)^2 = -4
(y-3)^2 - (x+2)^2/4 = 1
Now use that with your standard properties of hyperbolas.
check with this:
http://www.wolframalpha.com/input/?i=hyperbola+(y-3)%5E2+-+(x%2B2)%5E2%2F4+%3D+1
x^2+4x - 4(y^2-6y) = 28
x^2+4x+4 - 4(y^2-6y+9) = 28+4-4*9
(x+2)^2 - 4(y-3)^2 = -4
(y-3)^2 - (x+2)^2/4 = 1
Now use that with your standard properties of hyperbolas.
check with this:
http://www.wolframalpha.com/input/?i=hyperbola+(y-3)%5E2+-+(x%2B2)%5E2%2F4+%3D+1
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